laura venieri march 9, 2021 arxiv:1808.10233v1 [math.ca
TRANSCRIPT
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Geometric properties of extremal sets for the dimension
comparison in Carnot groups of step 2
Laura Venieri
March 9, 2021
Abstract
In Carnot groups of step 2 we consider sets having maximal or minimal possible
homogeneous Hausdorff dimension compared to their Euclidean one: in the first case
we prove that they must be in a sense vertical, that is a large part of these sets can
lie off horizontal planes through their points, and in the latter case close to horizontal.
The examples that had been used to prove the sharpness of the sub-Riemannian versus
Euclidean dimension comparison intuitively satisfied these geometric properties, which we
here quantify and prove to be necessary. We also show the sharpness of our statements
with some examples. This paper generalizes the analogue results proved by Mattila and
the author in the first Heisenberg group.
1 Introduction
Carnot groups are simply connected nilpotent Lie groups whose Lie algebra admits certaingradings, called stratifications. The first layer of the stratification is called horizontal and playsa special role since it generates all the Lie algebra by commutations. Using the stratification,one can define a metric, called Carnot-Carathéodory metric, which is of sub-Riemannian type,and a natural family of dilations. The Carnot-Carathéodory metric is bi-Lipschitz equivalentto any metric that is left-invariant with respect to the group operation and one-homogeneouswith respect to the dilations. We call such a metric homogeneous.
Carnot groups appear in several areas of mathematics, such as harmonic analysis, wherethey play a key role in the study of hypoelliptic differential operators (see e.g. [RS] and [S]).Carnot groups are metric tangents of sub-Riemannian manifolds, which are manifolds endowedwith a linear subbundle of the tangent bundle and an inner product on it, and have for exampleapplications in physics (see [Mo]) and in modelling the visual cortex (see [CS]).
Any Carnot group G is homeomorphic to Rn for a certain n, which means that topologically
they are the same. The Euclidean metric, however, is not bi-Lipschitz equivalent to anyhomogeneous metric, thus, for instance, the Hausdorff dimension of a subset of G in the
Key words. Hausdorff measure, Carnot groups, Hausdorff dimension, dimension comparison.
Mathematics Subject Classification. 28A75.
The author was supported by the Academy of Finland through the Finnish Center of Excellence in Analysis
and Dynamics research and through the project Quantitative rectifiability in Euclidean and non-Euclidean
spaces, grant no. 314172.
1
homogeneous and in the Euclidean metric can differ. We denote by dimG and dimE theHausdorff dimension in the homogenous and Euclidean sense respectively. In [BTW] (Theorem2.4) Balogh, Tyson and Warhurst solved the dimension comparison problem, determiningexplicit functions β± such that for every A ⊂ G,
β−(dimE A) ≤ dimGA ≤ β+(dimE A).
They also provide examples to show the sharpness of these inequalities, developing a theoryof fractal geometry in Carnot groups. Intuitively sets with dimGA = β+(dimE A) are asvertical as possible, that is they lie in the direction of the higher strata of the Lie algebra.Sets with dimGA = β−(dimE A), instead, lie in the direction of the lower strata, that is theyare as horizontal as possible. The tecnique used for constructing examples in the latter caseworked only for a dense set of dimensions and only for those the examples have positive andfinite Hausdorff measures at the respective dimensions. Later, Rajala and Vilppolainen [RV]constructed examples of such sets having positive and finite measures for any dimension,replacing self-similar constructions with so-called controlled Moran constructions.
This paper is a sequel of [MV], where Mattila and the author gave a characterization ofextremal sets for the dimension comparison in the first Heisenberg group H
1. We now proveanalogue results in Carnot groups of step 2, which have a stratification consisting of twolayers. In this case the functions β± have the form
β−(s) = max{s, 2s −m1}, β+(s) = min{2s, s +m2}, (1)
where m1 is the dimension of the first layer of the group (called horizontal) and m2 of thesecond (see Section 2.1 for more details). The Heisenberg group H
1, identified with R3, is the
simplest example of a Carnot group of step 2, where m1 = 2 and m2 = 1.Theorem 1 states that when 0 < s ≤ m1, which implies β−(s) = s, sets having positive
and finite Euclidean and homogeneous s-Hausdorff measures lie close to horizontal m1-planesin some arbitrarily small neighbourhoods of almost every point. Horizontal m1-planes, whichare obtained by left-translating the horizontal layer, are defined in (6). We show in Examples4, 5 that this result is sharp at least when s = 1 in the sense that we cannot take all smallneighbourhoods. Examples 6 and 7 show the sharpness in the case of Heisenberg groups H
m,identified with R
2m+1, when s is half the dimension of the first layer m1 = 2m. Example8, which relies on the construction in [RV], shows that when s ≥ m1 the same result doesnot hold since a large portion of the set can lie off horizontal planes through its points. InTheorems 2 and 3 we consider the case of sets with positive and finite s-dimensional EuclideanHausdorff measure and β+(s)-dimensional homogeneous Hausdorff measure: they cannot lietoo close to horizontal planes in small neighbourhoods of their typical points. Even if theproofs of the theorems are an adaptation of those in [MV], the results are of independentinterest and we explain in Section 5 that the result corresponding to Theorem 3 holds also ingeneral Carnot groups.
1.1 Acknowledgements
I would like to thank Pertti Mattila for many fruitful discussions about the topic.
2
2 Preliminaries
2.1 Carnot groups of step 2
A Carnot group of step 2 is a connected, simply connected Lie group whose Lie algebra g
admits a stratification of step 2, that is there exist linear subspaces V1, V2 such that
g = V1 ⊕ V2, [V1, V1] = V2, [V1, V2] = {0},
where [V1, Vi] is spanned by the commutators [X,Y ] with X ∈ V1, Y ∈ Vi. Let m1, m2 bethe dimensions of V1, V2 respectively.
Using exponential coordinates, we can identify G with Rn = R
m1 × Rm2 and denote
points by p = [p1, p2] = (x1, . . . , xm1, xm1+1, . . . , xn), where pi ∈ R
mi , i = 1, 2, and xj ∈ R,j = 1, . . . , n. The group operation has the form
[p1, p2] · [q1, q2] = [p1 + q1, p2 + q2 + P (p, q)],
where P = (Pm1+1, . . . , Pn) and each Pj is a polynomial of the form
Pj(p, q) = Pj(p1, q1) =
∑
1≤l<i≤m1
bjl,i(xlyi − xiyl), (2)
for some bjl,i ∈ R (see Lemma 1.7.2 in [Mon]). There is a natural family of dilations on G
given byδλ(p) = [λp1, λ2p2], λ > 0,
and each polynomial Pj is homogeneous of degree 2 with respect to δλ, e.g Pj(δλ(p), δλ(q)) =λ2Pj(p, q). The simplest examples of Carnot groups of step 2 are the Heisenberg groups H
m,m ≥ 1, identified with R
2m+1 = Rn, for which m1 = 2m and m2 = 1. Denoting the points by
p = (x, y, t) with x, y ∈ Rm, t ∈ R, the group operation takes the form
(x, y, t) · (x′, y′, t′) = (x+ x′, y + y′, t+ t′ + 2(〈y, x′〉 − 〈x, y′〉)), (3)
where 〈·, ·〉 denotes the scalar product.On Carnot groups one can define the Carnot-Carathéodory metric using horizontal curves,
which are tangent at almost every point to the vector fields generating V1 (see e.g. Section2.2 in [BTW]). Here we will work instead with the metric
d∞(p, q) = max{|p1 − q1|Rm1 , c|p2 − q2 + P (p1, q1)|1/2Rm2
}, (4)
where c ∈ (0, 1) is a suitable constant depending only on the group structure (see Theorem 5.1in [FSSC]). Here and in the following we will use | · | to denote the Euclidean norm and dropthe subscripts R
mi . Like the Carnot-Carathéodory metric, d∞ is left-invariant with respectto the group operation and one-homogeneous with respect to the dilations, i.e.
d∞(p · q, p · q′) = d∞(q, q′), d∞(δλ(p), δλ(q)) = λd∞(p, q)
for every p, q, q′ ∈ G and λ > 0. We call such metrics homogeneous and by Proposition 5.1.4in [BLU] they are all bi-Lipschitz equivalent. We denote by BE(p, r) and by B∞(p, r) the
3
closed ball of center p and radius r in the Euclidean metric dE and in the homogeneous metricd∞ respectively. Proposition 5.15.1 in [BLU] states that for every 0 < R < ∞ there existscR > 0 such that for every p, q ∈ BE(0, R) we have
1
cRdE(p, q) ≤ d∞(p, q) ≤ cRdE(p, q)
1/2. (5)
The horizontal m1-plane V (p) passing through a point p ∈ G is the left translation by pof {[q1, 0] : q1 ∈ R
m1}. It is the set of points q = [q1, q2] ∈ G such that
q2 = p2 + P (p1, q1). (6)
Note that if q ∈ V (p) thend∞(p, q) = |p1 − q1| ≤ dE(p, q) (7)
We will denote by V (p)(δ) the δ-neighbourhood of V (p) in the Euclidean metric, that is theset of points q ∈ G whose Euclidean distance from V (p) is ≤ δ. Note that if p ∈ BE(0, R)and 0 < r < 1, then
V (p)
(
r2
c2R
)
∩BE(p, r) ⊂ B∞(p, r) ⊂ V (p)
(
r2
c2
)
∩BE(p, cRr). (8)
To see that the right-hand side inclusion holds, let q = [q1, q2] ∈ B∞(p, r). Then dE(p, q) ≤cRd∞(p, q) ≤ cRr and
dE(q, V (p)) = infs∈V (p)
dE(q, s) =
= mins1∈Rm1
(|q1 − s1|2 + |q2 − p2 − P (p1, s1)|2)1/2, (9)
hence taking s1 = q1,
dE(q, V (p)) ≤ |q2 − p2 − P (p1, q1)| ≤ d∞(p, q)2
c2≤ r2
c2, (10)
which proves the right-hand side inclusion. For the other inclusion, let q = [q1, q2] ∈V (p)(r2/c2R)∩BE(p, r) and let q̄ = [p1, q2−P (p1, q1)]. Let s = [s1, p2+P (p1, s1)] ∈ V (p) be thepoint such that dE(q, V (p)) = dE(q, s) ≤ r2/c2R and let s̄ = [s1+p1− q1, p2+P (p1, s1− q1)] ∈V (p). We have
dE(q̄, s̄) =(
|s1 − q1|2 + |q2 − P (p1, q1)− p2 − P (p1, s1 − q1)|2)1/2
= dE(q, s) ≤r2
c2R, (11)
which implies that
d∞(q̄, s̄) ≤ cRdE(q̄, s̄)1/2 ≤ r.
On the other hand,
d∞(q̄, s̄) = max{|p1 − s1 − p1 + q1|, c|q2 − P (p1, q1)− p2 − P (p1, s1 − q1) + P (p1, s1 + p1 − q1)|1/2}= max{|q1 − s1|, c|q2 − p2 − P (p1, q1)|1/2},
4
hencec|q2 − p2 − P (p1, q1)|1/2 ≤ r.
This and the fact that q ∈ BE(p, r) imply that
d∞(p, q) = max{|p1 − q1|, c|p2 − q2 + P (p1, q1)|1/2} ≤ r,
which completes the proof of (8).
2.2 Hausdorff measures and dimensions
Given a metric space (X, d) and a real number 0 < s < ∞, the s-dimensional Hausdorffmeasure of a set A ⊂ X is defined as
Hs(A) = limδ→0
Hsδ(A),
where
Hsδ(A) = inf
∞∑
j=1
diamd(Aj)s : A ⊂
∞⋃
j=1
Aj ,diamd(Aj) < δ
and diamd denotes the diameter with respect to the metric d. The Hausdorff dimension of Awith respect to d is then defined by
dimd(A) = sup{s : Hs(A) > 0}.
We will denote by HsE the s-dimensional Hausdorff measure with respect to the Euclidean
metric, by HsG
the s-dimensional Hausdorff measure with respect to the metric d∞ and bydimE and dimG the corresponding Hausdorff dimensions. Note that the Hausdorff dimensionof a set with respect to any homogeneous metric is the same since the metrics are bi-Lipschitzequivalent.
We recall that the upper density theorem for Hausdorff measures, see for example 2.10.19in [F], states that for every Hs measurable set A ⊂ X with Hs(A) > 0, for Hs almost everyp ∈ A,
2−s ≤ lim supr→0
Hs(A ∩Bd(p, r))
(2r)s≤ 1 (12)
and for Hs almost every p ∈ X \ A,
limr→0
Hs(A ∩Bd(p, r))
(2r)s= 0. (13)
As explained in the introduction, Balogh, Tyson and Warhurst solved the dimension com-parison problem in Carnot groups, proving that for every A ⊂ G,
β−(dimE A) ≤ dimGA ≤ β+(dimE A),
where the functions β± were defined in (1) for a Carnot group of step 2. This result followsfrom Proposition 3.1 in the same paper [BTW] stating that for any 0 < R < ∞ there existsCR such that for any A ⊂ BE(0, R) and 0 ≤ s ≤ n,
Hβ+(s)G
(A)
CR≤ Hs
E(A) ≤ CRHβ−(s)G
(A). (14)
5
3 Results
We will now state and prove the theorems, whose content is a generalization to Carnotgroups of step 2 of those proved in [MV] for the first Heisenberg group H
1.
Theorem 1. Let 0 < s ≤ m1 and let A ⊂ G be such that HsG(A) < ∞. Then for Hs
E almostevery p ∈ A there exists 0 < ǫ < 1 such that
lim infr→0
HsE(A ∩BE(p, r) \ V (p)(r1+ǫ))
rs= 0.
Proof. Since the proof proceeds essentially as that of Theorem 2 in [MV], we recall here themain steps, proving in detail only the parts that need to be changed. We can assume that Ais a Borel set and that A ⊂ BE(0, R) for some R > 0. Using (14) we can reduce to the casewhen there is a constant C > 0 such that
1
CHs
E(B) ≤ HsG(B) ≤ CHs
E(B) (15)
for every B ⊂ A. The proof proceeds by contradiction, assuming that for ǫ > 0 there existc′ > 0, 0 < r0 < 1 and A′ ⊂ A with Hs
E(A′) > 0, such that
HsE(A ∩BE(p, r) \ V (p)(Kr1+ǫ)) > c′rs (16)
for every p ∈ A′ and 0 < r < r0, where K = 4/c2 + 1. We will get a contradiction at the endof the proof letting ǫ be small enough depending only on s, R and C. We can also assumeusing (12) that
HsE(A ∩BE(p, r)) ≤ 3srs and Hs
G(A′ ∩BH(p, r)) > rs/(2C) (17)
for r small enough so that r2 << r1+ǫ. Let k ∈ N be such that r(1+ǫ)k+1 ≤ r2 < r(1+ǫ)k .Using the 5r-covering theorem and (17), for j = 1, . . . , k we can find pj,i ∈ A′ ∩B∞(p, r) suchthat
A′ ∩B∞(p, r) =
Mj⋃
i=1
A′ ∩B∞(p, r) ∩BE(pj,i, 5r(1+ǫ)j ),
where the balls BE(pj,i, r(1+ǫ)j ), i = 1, . . . ,Mj , are disjoint and Mj ≥ rs(1−(1+ǫ)j)/(15s2C)
(here we use (17)).Now we show that the sets
mj⋃
i=1
BE(pj,i, r(1+ǫ)j ) \ V (pj,i)(Kr(1+ǫ)j+1
), j = 1, . . . , k, (18)
are disjoint. We show the details since they are a bit different from the case of H1. Let
j ∈ {1, . . . , k − 1} and let BE(pj,i, r(1+ǫ)j ) and BE(pm,l, r
(1+ǫ)m), j + 1 ≤ m ≤ k, i ∈{1, . . . ,Mj}, l ∈ {1, . . . ,Mm}, be such that BE(pj,i, r
(1+ǫ)j ) ∩ BE(pm,l, r(1+ǫ)m) 6= ∅. We
want to show that
(BE(pj,i, r(1+ǫ)j ) \ V (pj,i)(Kr(1+ǫ)j+1
)) ∩BE(pm,l, r(1+ǫ)m) = ∅. (19)
6
By the same calculation used to obtain (10) we have
dE(pm,l, V (pj,i)) ≤d∞(pm,l, pj,i)
2
c2≤(
d∞(pm,l, p) + d∞(p, pj,i)
c
)2
≤ 4r2
c2,
which implies that for every q ∈ BE(pm,l, r(1+ǫ)m)
dE(q, V (pj,i)) ≤4r2
c2+ r(1+ǫ)m ≤ Kr(1+ǫ)j+1
since r2 < r(1+ǫ)k ≤ r(1+ǫ)j+1
. Thus (19) holds. Then, as in the case of H1, we get by (16)
HsE(A ∩BE(p, 3r)) ≥
k∑
j=1
Mj∑
i=1
HsE(A ∩BE(pj,i, r
(1+ǫ)j ) \ V (pj,i)(Kr(1+ǫ)j+1
))
≥ c′k∑
j=1
Mjrs(1+ǫ)j ≥ 1
15s2C
k∑
j=1
rs(1−(1+ǫ)j)rs(1+ǫ)j ≥ k
15s2Crs.
Since k ≥ log 2/(2 log(1 + ǫ)), this is a contradiction with (12) when ǫ is small enough.
Theorem 2. Let s ≥ m2 and A ⊂ G be such that HsE(A) < ∞. Then for Hs+m2
Galmost
every p ∈ A there exists δ > 0 such that
lim supr→0
HsE(A ∩BE(p, r) \ V (p)(δr))
(2r)s>
1
2s+1. (20)
Proof. We follow essentially the proof of Theorem 3 in [MV], recall the main steps and showthe details to prove (24) since they are different. We can reduce to the case when A ∈ BE(0, R)is a Borel set and there is C > 0 such that
1
CHs
E(B) ≤ Hs+m2
G(B) ≤ CHs
E(B) (21)
for every B ⊂ A. Moreover, we can assume that there exists 0 < r0 < 1 such that for every0 < r < r0 and almost every p ∈ A,
HsE(A ∩BE(p, r)) ≤ 3srs and Hs+m2
G(A ∩B∞(p, r)) ≤ 3s+m2rs+m2 . (22)
More precisely, we would need to split A into a countable union of sets on which r0, whichdepends on the point p, is essentially 2−j , j ∈ N, but we skip these details. Let
0 < δ <c2
c2R(2s+5m2+13s+m2C)1/m2
(23)
and note that it depends only on s, R and C.We now show that for every 0 < r < min{r0, 4(cR/c)2δ} there exist p1, . . . , pk, k ≈
(δ/r)m2 , such that
BE(p, r) ∩ V (p)(δr) ⊂k⋃
i=1
B∞(pi, 2r). (24)
7
To prove it, let L(p) = {[q1, q2] ∈ G : q1 = p1}. If q ∈ L(p) and dE(q, V (p)) ≤ δr, then|q2 − p2| ≤ (cR/c)
2δr. To see this, note that there exists s = [s1, p2 + P (p1, s1)] ∈ V (p) suchthat dE(q, s) ≤ δr. This implies by (5) that d∞(q, s) ≤ cR(δr)
1/2 and since
d∞(q, s) =max{|p1 − s1|, c|q2 − p2 − P (p1, s1) + P (p1, s1)|1/2}= max{|p1 − s1|, c|q2 − p2|1/2},
we have c|q2 − p2|1/2 ≤ cR(δr)1/2, that is |q2 − p2| ≤ (cR/c)
2δr. Let us cover the Euclideanball with center p2 and radius (cR/c)
2δr in L(p) ⊂ Rm2 with Euclidean balls
BE(p2j , r
2), j = 1, . . . , k, k ≤(
2cRc
)2m2(
δr
r2
)m2
(25)
so that the corresponding balls of radii r2/2 are disjoint. Let pj = [p1, p2j ] ∈ L(p). To see
that (24) holds, let q = [q1, q2] ∈ BE(p, r) ∩ V (p)(δr) and let q̄ = [p1, q2 − P (p1, q1)] ∈ L(p).Then, proceeding as in the proof of (11), we also have dE(q̄, V (p)) ≤ δr.
Since q̄ ∈ L(p) ∩ V (p)(δr), which is contained in the Euclidean ball with center p2 andradius (cR/c)
2δr in L(p), there exists j ∈ {1, . . . k} such that dE(q̄, pj) = |q2−P (p1, q1)−p2j | ≤r2. Hence
d∞(pj , q) = max{|p1 − q1|, c|p2j − q2 + P (p1, q1)|1/2} ≤ max{r, cr} ≤ r,
which means that q ∈ B∞(pj, r). Thus (24) holds.Hence by (24), (21), (22), (25) and (23) we have for every 0 < r < r0
HsE(A ∩BE(p, r) ∩ V (p)(δr)) ≤
k∑
i=1
HsE(A ∩B∞(pi, 2r))
≤ C
k∑
i=1
Hs+m2
G(A ∩B∞(pi, 2r))
≤ Ck3s+m22s+m2rs+m2
≤ C
(
2cRc
)2m2
3s+m22s+3m2
(
δ
r
)m2
rs+m2 <1
2rs.
Thus for HsE almost every p ∈ A there exists δ > 0 such that
lim supr→0
HsE(A ∩BE(p, r) ∩ V (p)(δr))
(2r)s<
1
2s+1,
which proves (20) by (12) and (13).
Theorem 3. Let 0 < s < m2 and A ⊂ G be such that HsE(A) < ∞. Then for H2s
Galmost
every p ∈ A there exists δ > 0 such that
lim supr→0
HsE(A ∩BE(p, r) \ V (p)(δr))
(2r)s> 0. (26)
8
Proof. The proof proceeds as in the case of H1 with small changes (see Theorem 4 in [MV]),but we show it here for completeness. We may assume that A is a Borel, A ⊂ BE(0, R) forsome R > 0 and there exists C > 0 such that
1
CHs
E(B) ≤ H2sG (B) ≤ CHs
E(B) (27)
for every B ⊂ A.The proof proceeds by contradiction. Assume that we can find a Borel set A0 ⊂ A such
that HsE(A0) > 0 and (26) fails for p ∈ A0 for every δ > 0. Fix δ > 0 and ǫ > 0, which will
be chosen sufficiently small at the end of the proof. Then there exist a Borel set A′ ⊂ A0 andr0 > 0 such that Hs
E(A′) > Hs
E(A0)/2 and for every p ∈ A′ and for every 0 < r < r0,
HsE(A ∩BE(p, r) \ V (p)(δr)) < ǫrs. (28)
Let 0 < η < min{r0, δ}. By (12) for HsE almost all p ∈ A′ there exists rp < η such that
rsp3s
≤ HsE(A
′ ∩BE(p, rp)) ≤ 3srsp. (29)
We apply Vitali’s covering theorem (see e.g. Theorem 2.8 in [M]) to the family {BE(p, rp) :p ∈ A′ such that rp exists} to find a subfamily of disjoint balls, {BE(pi, ri)}∞i=1, such that
HsE
(
A′ \∞⋃
i=1
BE(pi, ri)
)
= 0. (30)
This implies by (29) that
HsE(A
′) = HsE
(
A′ ∩∞⋃
i=1
BE(pi, ri)
)
=
∞∑
i=1
HsE(A
′ ∩BE(pi, ri)) ≥1
3s
∞∑
i=1
rsi . (31)
Since pi ∈ BE(0, R), we have by (5) that
diam∞(BE(pi, ri)) ≤ cRdiamE(BE(pi, ri))1/2 ≤ cR(2η)
1/2. (32)
Let η′ = cR(2η)1/2. Then
H2sG,η′(A
′) ≤∞∑
i=1
H2sG,η′(A
′ ∩BE(pi, ri))
≤∞∑
i=1
H2sG,η′(A
′ ∩BE(pi, ri) ∩ V (pi)(δri)) (33)
+
∞∑
i=1
H2sG,η′(A
′ ∩BE(pi, ri) \ V (pi)(δri)).
Moreover, we claim that
H2sG,η′(A
′ ∩BE(pi, ri) ∩ V (pi)(δri)) ≤ diam∞(BE(pi, ri) ∩ V (pi)(δri))2s
≤ (2(cR + 2)(δri)1/2)2s = C ′′(δri)
s. (34)
9
To see this, let q, q′ ∈ BE(pi, ri) ∩ V (pi)(δri) and let q̄, q̄′ ∈ V (pi) be such that dE(q, q̄) ≤ δriand dE(q
′, q̄′) ≤ δri. Then
d∞(q, q′) ≤ d∞(q, q̄) + d∞(q̄, q̄′) + d∞(q̄′, q′).
We have by (5)
d∞(q, q̄) ≤ cRdE(q, q̄)1/2 ≤ cR(δri)
1/2, d∞(q′, q̄′) ≤ cRdE(q′, q̄′)1/2 ≤ cR(δri)
1/2 (35)
and by (7), since q̄, q̄′ ∈ V (pi),
d∞(q̄, q̄′) ≤ d∞(q̄, pi) + d∞(pi, q̄′) ≤ dE(q̄, pi) + dE(pi, q̄
′) ≤ 4ri.
Since ri < η < δ, it follows that r2i < δri, which implies ri ≤ (δri)1/2. Hence
d∞(q, q′) ≤ 2cR(δri)1/2 + 4ri ≤ 2(cR + 2)(δri)
1/2,
which proves (34). On the other hand, by (27) and (28) we have
H2sG (A′ ∩BE(pi, ri) \ V (pi)(δri)) ≤ CHs
E(A′ ∩BE(pi, ri) \ V (pi)(δri)) < Cǫrsi ,
thus alsoH2s
G,η′(A′ ∩BE(pi, ri) \ V (pi)(δri)) < Cǫrsi . (36)
Hence (33), (34), (36) and (31) imply that
H2sG,η′(A
′) ≤ (C ′′δs + Cǫ)
∞∑
i=1
rsi ≤ 3s(C ′′δs + Cǫ)HsE(A
′) ≤ 3s2(C ′′δs + Cǫ)HsE(A0).
Letting η (thus also η′) go to 0,
0 < H2sG (A0) < 2H2s
G (A′) ≤ 3s2(C ′′δs + Cǫ)HsE(A0).
Choosing δ and ǫ arbitrarily small, we get a contradiction, which completes the proof.
4 Examples
We now present some examples that show in which sense Theorem 1 is sharp. Examples4, 5, 6 and 7 show that the conclusion of Theorem 1 does not hold for any arbitrary smallneighbourhood. In particular, we cannot replace lim inf by lim sup or the r1+ǫ-neighbourhoodof V (p) by the r2-neighbourhood, which essentially behaves like the ball B∞(p, r) insideBE(p, r) by (8). The first two examples in Section 4.1 are 1-dimensional sets in any Carnotgroup of step 2, whereas the other two in Section 4.2 are m-dimensional sets in the Heisenberggroup H
m. Note that Hm can be identified with R
2m+1, where m1 = 2m, hence these setshave dimension m1/2. These four examples generalize those of H
1, see Examples 5 and 6in [MV]. Example 8 in Section 4.3 shows that in any Carnot group of step 2 sets of EuclideanHausdorff dimension s greater than m1 and homogeneous dimension β−(s) = 2s − m1 canhave positive density also far from horizontal planes.
10
4.1 Examples for s = 1
The following examples can be constructed in any Carnot group G of step 2 following thesame procedure used in H
1 for Examples 5 and 6 in [MV].
Example 4. There exists a compact set F ⊂ G such that for some positive constant C,H1
G(F ) > 0 and H1
G(A) ≤ CH1
E(A) < ∞ for A ⊂ F , and for p ∈ F ,
lim supr→0
H1E(F ∩BE(p, r) \ V (p)(r/8))
2r≥ 1
8.
Example 5. For any M, 1 < M < ∞, there exists a compact set F ⊂ G such that for somepositive constant C, H1
G(F ) > 0 and H1
G(A) ≤ CH1
E(A) < ∞ for A ⊂ F , and for p ∈ F ,
lim infr→0
H1E(F ∩BE(p, r) \ V (p)(Mr2))
2r≥ 1
16.
The sets in the examples will be subsets of the (x1, xm1+1)-plane
V1,m1+1 = {p = (x1, . . . , xm1, xm1+1, . . . , xn) ∈ G :
x2 = · · · = xm1= 0, xm1+2 = · · · = xn = 0}. (37)
Note that for p, q ∈ V1,m1+1, we have P (p1, q1) = 0 by (2) and we will denote the points byp = (x1, xm1+1), q = (y1, ym1+1). It follows that
d∞(p, q) = max{|x1 − y1|, c|xm1+1 − ym1+1|1/2}
andV (p) ∩ V1,m1+1 = {q = (y1, ym1+1) ∈ V1,m1+1 : ym1+1 = xm1+1}.
The sets in Examples 4 and 5 can now be constructed with the same procedure used in H1,
starting with the unit square [0, 1]2 ⊂ V1,m1+1, taking certaing subrectangles and iteratingthe construction. In the next section we will use the same procedure in higher dimensionsand so we do not repeat it here. In this case, since we are in the plane, it would be exactlythe same as the one used in [MV], which corresponds to m = 1 in the next section.
4.2 Examples for s = m1/2 in Heisenberg groups
We now construct similar examples to those in the previous section but having dimensionhalf the dimension of the first layer and only in Heisenberg groups. As briefly explained inSection 2.1, we denote points of Hm by p = (x, y, t) ∈ R
2m+1 and the group operation wasdefined in (3). A homogeneous metric that is commonly used in H
m is the Korányi one,defined by
dH(p, q) = ((|x− x′|2 + |y − y′|2)2 + (t− t′ + 2(〈x′, y〉 − 〈x, y′〉)2)1/4,
where q = (x′, y′, t′). We denote by HmH the m-dimensional Hausdorff measure with respect
to dH . The horizontal hyperplane V (p) passing through a point p = (x, y, t) consists of thosepoints q = (x′, y′, t′) ∈ H
m such that
t′ = t+ 2(〈x′, y〉 − 〈x, y′〉). (38)
11
Example 6. There exists a compact set F ⊂ Hm such that Hm
H(F ) > 0, HmH(A) ≤ CHm
E (A) <∞ for any A ⊂ F for some constant C > 0, and for p ∈ F ,
lim supr→0
HmE (F ∩BE(p, r) \ V (p)(r/8))
(2r)m≥(
1
8
)m
. (39)
Example 7. For any M , 1 < M < ∞, there exists a compact set F ⊂ Hm such that
HmH(F ) > 0, Hm
H(A) ≤ CHmE (A) < ∞ for A ⊂ F for some constant C > 0, and for p ∈ F ,
lim infr→0
HmE (F ∩BE(p, r) \ V (p)(Mr2))
(2r)m≥(
1
25
)m
. (40)
Both examples are based on the following construction, which is a simple generalizationof the one used in Examples 5 and 6 [MV]. Let
Vx,t = {(x, y, t) ∈ Hm : y = 0} ⊂ R
m+1.
In both examples the set F will be a subset of Vx,t. We denote the points of Vx,t by p = (x, t).For p = (x, t), q = (x′, t′) ∈ Vx,t we have
dH(p, q) = (|x− x′|4 + |t− t′|2)1/4.
For p = (x, t) ∈ Vx,t, (38) implies that
V (p) ∩ Vx,t = {(x′, t′) : t′ = t}.
Remark. The fact that V (p)∩Vx,t is parallel to the x-coordinates plane simplifies significantlythe construction of Examples 6 and 7. In a step 2 Carnot group G we cannot in general havethis simplification, since it would correspond to finding a certain V ′ ⊂ G such that
P (p1, q1) = 0 for all p, q ∈ V ′, (41)
which is what happens in Hm when V ′ = Vx,t. This holds, for example, when the points in
V ′ have all coordinates of the first layer equal to zero except for 1, as in the case of V1,m1+1
defined in (37). In this case, however, we would have V ′ ⊂ R1+m2 . We were not able to find
another easy way to ensure that (41) would hold if, for example, only half of the coordinatesof Rm1 were zero, as in the case of Vx,t in H
m. For this reason we construct the examplescorresponding to s = m1/2 only in H
m.
Given L > 0, an integer N ≥ 1, 0 < λ ≤ 12 and the (m+ 1)-dimensional parallelepiped
R = [a1, b1]× · · · × [am, bm]× [c, d] ⊂ Vx,t
such that bi − ai = L for every i = 1, . . . ,m and λL ≤ (d− c)/2, we choose (2N)m subsets ofR, which are (m+1)-dimensional parallelepipeds, in the following way. We divide
∏mi=1[ai, bi]
into (2N)m identical m-dimensional cubes Cj, j = 1, . . . , (2N)m, of side length L/2N . Wechoose R1 = Rc
1 = C1× [c, c+λL]. Then for each j we define either Rj = Rcj = Cj× [c, c+λL]
or Rj = Rdj = Cj × [d − λL, d] in such a way that two distinct Rc
i , Rcj do not share an m-
dimensional face (and the same for Rdi , R
dj ). Hence we have (2N)m/2 sub-parallelepipeds of
12
the form Rci and (2N)m/2 of the form Rd
i . Then we define R(R,N, λ) as the collection ofthese (2N)m parallelepipeds and we have that
πx
⋃
R∈R(R,N,λ)
R
=
m∏
i=1
[ai, bi], (42)
where πx(x, t) = x.The case m = 1 corresponds to the construction in Examples 5 and 6 in [MV]. Let us
write down explicitly the case m = 2. Given R = [a1, b1]× [a2, b2]× [c, d], with bi−ai = L, aninteger N ≥ 1 and 0 < λ ≤ 1
2 such that λL ≤ (d− c)/2, R(R,N, λ) is defined as the followingcollection of 4N2 sub-parallelepipeds of R.
• For i, j = 0, 1, . . . , N − 1, let
Rci,j =
[
a1 + 2iL
2N, a1 + 2i
L
2N+
L
2N
]
×[
a2 + 2jL
2N, a2 + 2j
L
2N+
L
2N
]
× [c, c+ λL]
and for l, k = 0, 1, . . . , N − 1,
Rcl,k =
[
a1 + (2l + 1)L
2N, a1 + (2l + 1)
L
2N+
L
2N
]
×[
a2 + (2k + 1)L
2N, a2 + (2k + 1)
L
2N+
L
2N
]
× [c, c+ λL].
Thus we have 2N2 sub-parallelepipeds of this form.
• For i, j = 0, 1, . . . , N − 1, let
Rdi,j =
[
a1 + 2iL
2N, a1 + 2i
L
2N+
L
2N
]
×[
a2 + (2j + 1)L
2N, a2 + (2j + 1)
L
2N+
L
2N
]
× [d− λL, d]
and for l, k = 0, 1, . . . , N − 1,
Rdl,k =
[
a1 + (2l + 1)L
2N, a1 + (2l + 1)
L
2N+
L
2N
]
×[
a2 + 2kL
2N, a2 + 2k
L
2N+
L
2N
]
× [d− λL, d].
Thus we have 2N2 sub-parallelepipeds of this form.
Let us now construct the sets in Examples 6 and 7 for any m ≥ 1. Given a sequence ofintegers (Nk)k, Nk ≥ 1, and a sequence of positive numbers (λk)k, λk ≤ 1/2, we define fork ≥ 1,
R0 = R([0, 1]m+1, 1, 1/2),
Rk =⋃
R∈Rk−1
R(R,Nk, λk),
13
and
F =
∞⋂
k=0
⋃
R∈Rk
R.
It follows that F ⊂ Vx,t is compact and πx(F ) = [0, 1]m by (42), thus both HmE (F ) and
HmH(F ) are at least 1. We can also check, using the natural coverings with the parallelepipeds
of Rk, that these measures are finite if λk goes to 0 sufficiently fast. To see this, let hk bethe length of the sides in the directions of the x-coordinates of the parallelepipeds of Rk (allsides in the directions of the x-coordinates have the same length) and let vk be the length oftheir vertical sides (in the t-direction). Note that
h0 =1
2, hk =
hk−1
2Nkfor k ≥ 1, v0 =
1
2, vk = λkhk−1 for k ≥ 1.
For every R ∈ Rk, we have
diamE(R) = (mh2k + v2k)1/2 and diamH(R) = (m2h4k + v2k)
1/4.
If vk/hk tends to zero as k → ∞, then for R ∈ Rk,
hmk ≤ HmE (F ∩R) ≤ (
√mhk)
m, (43)
which implies
1 ≤ HmE (F ) ≤ (#Rk)(
√mhk)
m = (2Nk)m(
√mhk)
m ≤ mm/2.
If moreover, vk ≤ Ch2k for some 0 < C < ∞ for all large enough k, then for A ⊂ F ,
HmH(A) ≤ (m2 + C2)1/4√
mHm
E (A). (44)
In Example 6 we will choose vk = h2k and in Example 7 vk = 136Mmh2k for large k, hence theabove conditions on hk and vk will be satisfied.
For Example 6 let Nk = 22k−1
−1
and λk = 2−3·2k−1
. It follows that hk = 2−2
k
and
vk = 2−2k+1
for k ≥ 1. Thus hk+1 = vk, which means that for R ∈ Rk, the horizontal sides ofeach parallelepiped R′ of Rk+1 inside R have the same length as the vertical sides of R. Thisimplies that for p ∈ R′ and rk = 4hk+1, BE(p, rk)\V (p)(rk/8) contains another parallelepipedR′′ of Rk+1. Indeed, if R′ is of the form Rc
i for some i then there is one Rdj that is at least
at vertical distance vk − 2vk+1 > vk/2 = rk/8 from p, so it is outside V (p)(rk/8), and iscontained in BE(p, rk) (and similarly if R′ is of the form Rd
i we can find R′′ of the form Rcj).
Hence, using (43),
HmE (F ∩BE(p, rk) \ V (p)(rk/8))
(2rk)m≥ Hm
E (F ∩R′′)
(8hk+1)m=
(
1
8
)m
,
from which (39) follows. Recalling also (44), we have completed Example 6.For Example 7, given 1 < M < ∞, we let Nk = 1 for all k ≥ 1 and λk = 1/2 when
68Mm4−k ≥ 2−k, that is, 2k ≤ 68Mm, and λk = 68Mm2−k−1 for all larger k. Thus,for large enough k so that 2k > 68Mm, we have hk = 2−k−1 and vk = 34Mm4−k. For
14
R ∈ Rk, there are 2m sub-parallelepipeds Rj of Rk+1 inside R, of which half are of theform Rd
j and half of the form Rci . The distance between any Rd
j and any Rci is at least
vk − 2vk+1 = 34Mm4−k − 2 · 34Mm4−k−1 = 17Mm4−k.Given 0 < r < 1, let k be such that
√m21−k ≤ r <
√m22−k. We may assume that r is
small enough so that 2k > 68Mm. Let R,Rdj and Rc
i be as in the previous paragraph and p ∈Rd
j . Since Mr2 ≤ Mm42−k < 17Mm4−k, all the parallelepipeds Rci ’s lie outside V (p)(Mr2).
On the other hand, note that√mhk =
√m2−k−1 ≤ r/2 and vk = 34Mm4−k ≤ r/2 because
2k > 68Mm. Hence we have
diamE(R) ≤√
mh2k + v2k ≤√
r2
4+
r2
4≤ r,
which means that each Rci lies inside BE(p, r). This implies that BE(p, r)\V (p)(Mr2) contains
(at least one) Rci , thus
HmE (F ∩BE(p, r) \ V (p)(Mr2))
(2r)m≥ Hm
E (F ∩Rci )
(2r)m≥(√
m2−k−2
√m23−k
)m
=
(
1
25
)m
,
which is what we wanted to show in Example 7.
4.3 Example for m1 ≤ s ≤ n
The last example that we present shows that in any Carnot group G of step 2 the conclusionof Theorem 1 does not hold when s ≥ m1.
We will use the following estimate. For any p1 = (x1, . . . , xm1), q1 = (y1, . . . , ym1
),
|P (p1, q1)|2 =n∑
j=m1+1
Pj(p1, q1)2
=
n∑
j=m1+1
∑
1≤l<i≤m1
bjl,i(xlyi − xiyl)
2
=
n∑
j=m1+1
∑
1≤l<i≤m1
bjl,i〈(xl, yl), (yi,−xi)〉
2
(45)
≤ m2m1(m1 − 1) maxm1+1≤j≤n
max1≤l<i≤m1
(bjl,i)2|p1 − q1|2|p1 + q1|2
≤ C2R|p1 − q1|2,
where after (45) we used Cauchy-Schwarz and
CR = 2R
√
m2m1(m1 − 1) maxm1+1≤j≤n
max1≤l<i≤m1
(bjl,i)2. (46)
15
Instead of line (45) we can also do the following.
|P (p1, q1)|2 =n∑
j=m1+1
∑
1≤l<i≤m1
bjl,i〈(xl,−xi), (yi, yl)〉
2
≤ m2m1(m1 − 1) maxm1+1≤j≤n
max1≤l<i≤m1
(bjl,i)2|p1|2|q1|2
≤ C2R|q1|2. (47)
Example 8. For any m1 ≤ s ≤ n, there exist cs, δs > 0 and a compact set Fs ⊂ G such thatHs
E(Fs) > 0, H2s−m1
G(Fs) < ∞ and for Hs
E-almost every p ∈ Fs,
lim infr→0
HsE(Fs ∩BE(p, r) \ V (p)(δsr))
rs≥ cs. (48)
This construction comes from Thereom 5.4 in [RV], where it is done in Carnot groupsof any step, and it modifies the self-similar construction used in Proposition 4.14 in [BTW].We use the same notation, which we recall here. Let m1 ≤ s ≤ n. Let A1 = {0, 1}m1 ,A2 = {0, 1, 2, 3}m2 and define
F1 = {Fa1 : a1 ∈ A1}and
F2 = {Fa1a2 : a1 ∈ A1, a2 ∈ A2},where
Fa1(p) = [a1, 0] · δ1/2([−a1, 0] · p)and
Fa1a2(p) = [a1, a2] · δ1/2([−a1,−a2] · p).Then define a sequence (ni)i∈N by induction as follows. Let n1 = 2 and, assuming thatn1, . . . , nt have been defined, let nt+1 = 2 if
t∏
i=1
n2m2
i < 22t(s−m1)
and nt+1 = 1 otherwise. If ni = 1 we will use the system F1 at step i, otherwise we willuse F2. Write F2 = {g1, g2, . . . , g2m1+2m2}, where gi /∈ F1 for 2m1 < i ≤ 2m1+2m2 . LetI = {1, 2, . . . , 2m1+2m2}, In = {i = (i1, . . . , in) : ij ∈ I}, I∗ = ∪∞
n=1In, I∞ = IN and
J = {i = (i1, i2, . . . ) : ij ∈ N, 1 ≤ ij ≤ n2m2
j 2m1} ⊂ I∞.
Then we write Jn = {i ∈ In : [i] ∩ J 6= ∅} and J∗ = ∪∞n=1Jn. Here [i] is the cylinder
set of i, that is the set of ij ∈ I∞ obtained by juxtaposing i ∈ In and j ∈ I∗ ∪ I∞. Fori = (i1, . . . , it) ∈ It let
Xi = gi1 ◦ · · · ◦ git(E),
where E is the attractor of F2. For i = (i1, . . . , in) ∈ In we denote by |i| = n its length,by i|m ∈ Im with 1 ≤ m < n the element (i1, . . . , im) and by i− the element i|n−1. Thecollection {Xi : i ∈ J∗} is a (2s−m1)-controlled Moran sub-construction, which means thatit satisfies the following conditions:
16
(i) Xi ⊂ Xi− ,
(ii) there exist a constant D ≥ 1 and n ∈ N such that
maxi∈Jn
diam∞(Xi) < 1/D,
(iii) there exists a constant C > 0 such that for every i ∈ J∗ and m ∈ N,
diam∞(Xi)2s−m1
C≤
∑
j∈Im,ij∈J∗
diam∞(Xij)2s−m1 ≤ Cdiam∞(Xi)
2s−m1 . (49)
We refer to Section 5 in [RV] for more details and examples of Moran sub-constructions. Sincethe maps gj are all contractions of ratio 1/2 with respect to d∞, we have
diam∞(Xi) ≤1
2|i|diam∞(E).
Moreover, in (49) we can take C = 24m2 . Let Fs be the limit set of this sub-construction. Itis shown in Theorem 5.4 in [RV] that Fs is compact,
H2s−m1
G(Fs) < ∞ and Hs
E(Fs) > 0.
Let π1 : G → Rm1 and π2 : G → R
m2 be the orthogonal projections to the first and secondlayer respectively. Note that for every a1 ∈ A1, a2 ∈ A2 and p = [p1, p2] ∈ G,
π1(Fa1([p1, p2]) =
1
2(p1 − a1) + a1 = π1(Fa1a2([p
1, p2]),
which means that π1(Fs) is the invariant set for the Euclidean self-similar iterated functionsystem {π1(Fa1) : a1 ∈ A1}. The construction used in Proposition 4.14 in [BTW] gives thesame. It follows that π1(Fs) is a Euclidean self-similar set. It also satisfies the open setcondition with the open set (0, 1)m1 , thus Hm1
E (π1(Fs)) > 0 and it is m1-Ahlfors regular.Now for almost every q1 ∈ π1(Fs) the set π2(Fs ∩ π−1
1 (q1)) is a Euclidean translate of thelimit set K ′ of the Euclidean construction {Yi : i ∈ J ′
∗} with
Yi = hi1 ◦ · · · ◦ him([0, 2]m2),
J ′ = {i = (i1, i2, . . . ) : 1 ≤ ij ≤ n2m2
j },
hij (y) =1
4y +
3
4aij .
This is different from the construction in [BTW], where these level sets π2(Fs ∩ π−11 (q1)) are
Euclidean translates of an invariant set of a Euclidean self-similar iterated function system.The collection {Yi : i ∈ J ′
∗} is an (s−m1)-Moran sub-construction, so it satisfies properties(i), (ii), (iii) above with respect to the Euclidean metric. In particular, there is a constantC > 0 such that for every i ∈ J ′
∗ and n ∈ N,
diamE(Yi)s−m1
C≤
∑
j∈In,ij∈J ′
∗
diamE(Yij)s−m1 ≤ CdiamE(Yi)
s−m1 . (50)
17
We can take again C = 24m2 . In the proof of Theorem 5.4 in [RV] it is also shown thatHs−m1
E (K ′) > 0 since it satisfies the finite clustering property, which is a separation conditiondefined in Section 3 in [RV] stating the following. Letting for y ∈ K ′ and r > 0,
Z(y, r) = {i ∈ J ′∗ : diamE(Yi) ≤ r < diamE(Yi−), Yi ∩BE(y, r) 6= ∅},
we haveZ = sup
y∈K ′
lim supr→0
#Z(y, r) < ∞.
Writing K ′ = ∪∞l=1K
′l , where K ′
l = {y ∈ K ′ : #Z(y, r) ≤ Z ∀ 0 < r < 1/l}, we could run theargument below for every l such that Hs−m1
E (K ′l) > 0 so we can assume that #Z(y, r) ≤ Z
for every y ∈ K ′ and 0 < r < r0 for a certain r0 > 0.We claim that for every y ∈ K ′ and every 0 < r < r0 we have
rs−m1
Cs≤ Hs−m1
E (K ′ ∩BE(y, r)) ≤ Csrs−m1 , (51)
for some constant Cs > 0. To this end, for y ∈ K ′ and 0 < r < r0, note that for every m ∈ N
K ′ ∩BE(y, r) ⊂⋃
i∈Z(y,r)
⋃
j∈Im,ij∈J ′
∗
Yij,
hence by (50)
Hs−m1
E (K ′ ∩BE(y, r)) ≤∑
i∈Z(y,r)
∑
j∈Im,ij∈J ′
∗
diamE(Yij)s−m1
≤∑
i∈Z(y,r)
CdiamE(Yi)s−m1 ≤ ZCrs−m1,
which proves the right-hand side inequality in (51). Let i ∈ J ′ be such that y = π(i), whereπ is the projection mapping J ′ → K ′, {π(i)} = ∩∞
n=1Yi|n . Note that
diamE(Yi|n) =2√m2
4n.
Given 0 < r < r0, let n be such that
2√m2
4n≤ r ≤ 2
√m2
4n−1,
so that Yi|n ⊂ BE(y, r). Let ǫ =√m2/(C4n(s−m1)). There exists m > n sufficiently large so
that
Hs−m1
E (K ′ ∩BE(y, r)) ≥∑
j∈Im−n,i|nj∈J ′
m
diamE(Yi|nj)s−m1 − ǫ,
noting that the sets Yi|nj are disjoint. Using (50) it follows that
Hs−m1
E (K ′ ∩BE(y, r)) ≥diamE(Yi|n)s−m1
C− ǫ =
2√m2
C4n(s−m1)− 2
√m2
2C4n(s−m1)≥ rs−m1
Cs,
18
which completes the proof of (51).Let p = [p1, p2] ∈ Fs and
0 < r < min
{
r0,c2CR
40(2 + 5C2R)
,CRc
2c02
}
, (52)
where CR is the constant in (46) and c0 is defined below. Let Xi be such that p ∈ Xi
and diam∞(E)/2|i| ≤ r < diam∞(E)/2|i|−1. Then Xi ⊂ B∞(p, r). For almost every q =[q1, q2] ∈ Xi such that |q1 − p1| ≤ r/40(2 + 5C2
R), we have the following. By redefining K ′ asa translation of itself if needed, we can assume that π2(π
−11 (q1)∩Fs) = K ′+ q2 ⊂ R
m2 . Thenwe let
Lq = {[q1, q2 + t] ∈ Fs : t ∈ K ′, c0r ≤ |t| ≤ r/(2 + 5C2R)}
for some constant c0 such that
Hs−m1
E (π2(Lq)) ≥ C ′sr
s−m1 . (53)
The existence of c0 follows from the fact that π2(Lq) = (K ′ + q2) ∩ BE(q2, r/(2 + 5C2
R)) \BE(q
2, c0r) and from (51). Let
Dr(p) = Fs ∩BE(p, r) ∩{
[u1, u2] : |u1 − p1| ≤ r
40(2 + 5C2R)
}
\ V (p)
(
c0r
2√2CR
)
.
We claim that
Lq ⊂ Dr(p) for almost every q ∈ B∞(p, r) ∩ Fs, |q1 − p1| ≤ r
40(2 + 5C2R)
. (54)
Let qt = [q1, q2 + t] ∈ Lq. By (45), the fact that q ∈ Xi ⊂ B∞(p, r) and (52) we have
|q2 − p2| ≤ |q2 − p2 − P (p1, q1)|+ |P (p1, q1)|
≤ d∞(p, q)2
c2+ CR|p1 − q1| ≤ r2
c2+ CR
r
40(2 + 5C2R)
≤ CR
20(2 + 5C2R)
r.
Hence
dE(p, qt)2 = |p1 − q1|2 + |p2 − q2 − t|2 ≤ r2
402(2 + 5C2R)
2+
(
CR
20(2 + 5C2R)
r +r
2 + 5C2R
)2
≤ r2,
which implies that qt ∈ BE(p, r). Since qt ∈ Fs, it remains to show that dE(qt, V (p)) >c0r/(2
√2CR). As in (9), we have that
dE(qt, V (p)) = mins1∈Rm1
(|q1 − s1|2 + |q2 + t− p2 − P (p1, s1)|2)1/2.
For every s1 ∈ Rm1
|t| ≤ |q2 + t− p2 − P (p1, s1)|+ | − q2 + p2 + P (p1, s1)|≤ |q2 + t− p2 − P (p1, s1)|+ | − q2 + p2 + P (p1, q1)|+ |P (p1, s1)− P (p1, q1)|
≤ |q2 + t− p2 − P (p1, s1)|+ d∞(p, q)2
c2+ CR|s1 − q1|,
19
where in the last inequality we used (47) as follows:
|P (p1, s1)− P (p1, q1)| = |P (p1, s1 − q1)| ≤ CR|s1 − q1|.
Hence
dE(qt, V (p)) ≥ mins1
1√2CR
(CR|s1 − q1|+ |q2 + t− p2 + P (p1, s1)|)
≥ 1√2CR
(
|t| − d∞(p, q)2
c2
)
≥ 1√2CR
(
c0r −r2
c2
)
>c0
2√2CR
r,
where we used (52). This concludes the proof of (54). It follows that for almost everyq ∈ Xi with |q1 − p1| ≤ r/40(2 + 5C2
R) there exists a point [q1, q2 + t] ∈ Dr(p), so π1(q) =π1[q
1, q2 + t]) ∈ π1(Dr(p)). Hence
Hm1
E (π1(Dr(p))) ≥ Hm1
E (π1(Xi ∩ {u1 ∈ Rm1 : |u1 − p1| ≤ r/40(2 + 5C2
R)}))= Hm1
E (π1(Xi) ∩ {u1 ∈ Rm1 : |u1 − p1| ≤ r/40(2 + 5C2
R)}).
Using the fact that π1(Fs) is m1-Ahlfors regular, we have
Hm1
E (π1(Dr(p))) ≥ Crm1
for some constant C > 0. Hence by (53) and Theorem 7.7 in [M] we have for some constantc′ > 0,
HsE
(
Fs ∩BE(p, r) \ V (p)
(
c0r
2√2CR
))
≥ HsE(Dr(p))
≥ c′∫
π1(Dr(p))Hs−m1
E ({[0, q2] : [q1, q2] ∈ Dr(p)})dHm1
E (q1)
≥ c′∫
π1(Dr(p))Hs−m1
E (π2(Lq))dHm1
E (q1)
≥ c′C ′sr
s−m1Hm1
E (π1(Dr(p))) ≥ csrs−m1rm1 = csr
s,
which proves (48).
5 Some Remarks about general Carnot groups
We recall here few basic facts about general Carnot groups and show that the analogueof Theorem 3 always holds. Let G be a Carnot group of step k ≥ 2, which means that its Liealgebra g admits a stratification of the form
g = V1 ⊕ · · · ⊕ Vk, [V1, Vi] = Vi+1, Vk 6= {0}, Vi = {0}, i > k.
Using exponential coordinates, G can be identified with Rn = R
m1 × · · · × Rmk , where mi
is the dimension of Vi. We denote points by p = [p1, . . . , pk] with pj ∈ Rmj . The group
operation has the form
p · q = [p1 + q1, p2 + q2 + P2(p, q), . . . , pk + qk + Pk(p, q)],
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where Pj = (Pj,1, . . . , Pj,mj) and each Pj,i is a polynomial of degree j with respect to the
dilationsδλ(p) = [λp1, λ2p2, . . . , λkpk], λ > 0.
Moreover, for every p, q ∈ G,
Pj,i(p, 0) = Pj,i(0, q) = 0, Pj,i(p, p) = Pj,i(p,−p) = 0
andPj,i(p, q) = Pj,i([p
1, . . . , pj−1], [q1, . . . , qj−1]).
As in the case of step 2, we consider the following metric, which is bi-Lipschitz equivalentto the Carnot-Carathéodory one:
d∞(p, q) = max{|p1 − q1|, ǫ2|p2 − q2 + P2(p, q)|1/2, . . . , ǫk|pk − qk + Pk(p, q)|1/k},where ǫj ∈ (0, 1), j = 2, . . . k, are constants depending only on the group structure (seeTheorem 5.1 in [FSSC]). By Proposition 5.15.1 in [BLU] for every 0 < R < ∞ there existscR > 0 such that for every p, q ∈ BE(0, R) we have
1
cRdE(p, q) ≤ d∞(p, q) ≤ cRdE(p, q)
1/k. (55)
For a Carnot group of step k the functions β± appearing in the dimension comparisontheorem (see Theorem 2.4 in [BTW]) have the form
β−(s) =
s, 0 < s ≤ m1
2s−m1, m1 < s ≤ m1 +m2
. . .
ks− (k − 1)m1 − · · · −mk−1, m1 + . . . mk−1 < s ≤ n
(56)
and
β+(s) =
ks, 0 < s ≤ mk
(k − 1)s +mk, mk < s ≤ mk−1 +mk
. . .
s+m2 + 2m3 + · · · + (k − 1)ms, m2 + . . . mk < s ≤ n.
(57)
The local Hausdorff measure comparison (14) still holds with these functions β±. The hor-izontal m1-dimensional subspace V (p) passing through a point p is given by those pointsq = [q1, . . . , qk] ∈ G such that
qj = pj + Pj(p, q), j = 2, . . . , k. (58)
Note that if q ∈ V (p), then d∞(p, q) = |p1 − q1| ≤ dE(p, q).As you can see form (56) and (57), for general Carnot groups there would be many cases
to consider for the dimension comparison (double the step of the group) and we do not knowwhat kind of geometric properties extremal sets should satisfy in each case. It seems naturalto believe that also here when s ≤ m1 sets with positive and finite Euclidean and homogeneouss-Hausdorff measures must be in some sense horizontal but we do not know how to prove it.However, we can show that in the case when 0 < s < mk, which implies β+(s) = ks, sets withfinite s-dimensional Euclidean Hausdorff measure must be in some sense vertical. Indeed, thefollowing version of Theorem 3 holds.
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Theorem 9. Let 0 < s < mk and A ⊂ G be such that HsE(A) < ∞. Then for Hks
Galmost
every p ∈ A there exists δ > 0 such that
lim supr→0
HsE(A ∩BE(p, r) \ V (p)(δr))
(2r)s> 0. (59)
The proof is basically the same as in the case of step 2 groups with few small changes thatwe explain here. The first change is in equation (32), which becomes using (55),
diam∞(BE(pi, ri)) ≤ cRdiamE(BE(pi, ri))1/k ≤ cR(2η)
1/k .
Then, taking η′ = cR(2η)1/k , (34) becomes
HksG,η′(A
′ ∩BE(pi, ri) ∩ V (pi)(δri)) ≤ diam∞(BE(pi, ri) ∩ V (pi)(δri))ks
≤ (2(cR + 2)(δri)1/k)ks = C ′′(δri)
s. (60)
This can be proved in the same way as (34) where (35) becomes
d∞(q, q̄) ≤ cRdE(q, q̄)1/k ≤ cR(δri)
1/k, d∞(q′, q̄′) ≤ cRdE(q′, q̄′)1/k ≤ cR(δri)
1/k
and we use the fact that ri < δ implies ri ≤ (δri)1/k. The rest of the proof is exactly the
same as for step 2 groups.
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Department of Mathematics and Statistics, P.O. Box 68, FI-00014 University of Helsinki,
Finland,
E-mail addresses: [email protected]
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